How Does Positive Second Derivative Imply Function Convexity in Calculus?

[Amad27]

I recently searched around SE, and found:

[How to solve this derivative of f proof][1] [1]: calculus - How to solve this derivative of f proof? - Mathematics Stack Exchange

The answer is interesting.

"A function given that $f(x)=f''(x)+f'(x)g(x)$ could be an exponential function, sine, cosine , quadratic polynomial or $f\equiv0$. So we can say that the function is a continuous function $\in C^2$.

The right negation is that $f(x)\ge0$ in $(a,b)$ and exist a point c | $f(c)>0$.


You have that $f''(x_1)\ge0$ (the function in that point is convex) so in that point you have a minima so there are two case

1. $f(x_1)<0$ (obviously contradiction)
2. $f(x_1)=0$ (it's impossible because this imply that $f(x)=0$ $ \forall x \in (a,b) $)

Analog for the other case"

But how does $f''(x_1) > 0$ show that $f(x_1) < 0$?

Thanks!

 

[jvicens]

Thank you for bringing up this question. To answer your question, let's first recall the definition of convexity: a function $f(x)$ is convex on an interval if for any two points $x_1$ and $x_2$ in that interval, the function lies above the line connecting these two points. This can also be stated as $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda)f(x_2)$, where $\lambda$ is a value between 0 and 1.

Now, let's consider the point $x_1$ where $f''(x_1) > 0$. This means that the second derivative of $f(x)$ at this point is positive, which implies that the function is concave up. In other words, the function lies above the tangent line at this point. From the definition of convexity, we know that the function must also lie above the line connecting $x_1$ and any other point $x_2$ in the interval.

Now, if $f(x_1) < 0$, then the function would lie below the tangent line at $x_1$, which contradicts our assumption that the function is convex at this point. Therefore, we can conclude that $f(x_1) > 0$. I hope this helps to clarify the proof for you.